1.

Consider the reaction : N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g) The equality relationship between (d[NH_(3)])/(dt) and -(d[H_(2)])/(dt) is

Answer»

`+(d[NH_(3)])/(DT)=-(2)/(3) (d[H_(2)])/(dt)`
`+(d[NH_(3)])/(dt)= -(3)/(2) (d[H_(2)])/(dt)`
`(d[NH_(3)])/(dt)= -d(H_(2)])/(dt)`
`(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`

SOLUTION :(A) `-(1)/(2) (d[N_(2)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)=(1)/(2) (d[NH_(3)])/(dt)`
or `(d[NH_(3)])/(dt) =-(2)/(3)(d[H_(2)])/(dt)`


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